If $u''>0$ in $\mathbf{R}^+$ then $u$ is unbounded? If $u$ is a positive function such that $u''>0$ in the whole $\mathbf{R}^+$ then $u$ is unbounded?
In fact, I know that if $u''>0$ then $u$ is strictly convex. I think that implies $u$ is coercive. I want to prove it.
 A: $$ u(x) = e^{-x}  {}{}{}{}{}{}{}{} $$
EDIT: if you actually meant the entire real line $\mathbb R,$ then any $C^2$ function $u(x)$ really is unbounded. Proof: as $u'' > 0,$ we know that $u'$ cannot always be $0.$ as a result, it is nonzero at some $x=a.$ If $u'(a) > 0,$ then for $x > a$ we have $u(x) > u(a) + (x-a) u'(a),$ which is unbounded. If, instead, $u'(a) < 0,$ then for $x < a$ we have $u(x) > u(a) + (x-a) u'(a),$ which is unbounded as $(x-a)$ is negative. Both of these are the finite Taylor theorem. Examples with minimal growth include
$$  x + \sqrt{1 + x^2}   $$ and
$$  -x + \sqrt{1 + x^2}   $$ 
Note that $C^2$ is not required, it suffices that the second derivative always exist and is always positive. Taylor's with remainder.
A: Consider $g(x)=e^{-x^2}$. Observe that this is a bounded, positive, everywhere differentiable function, and that for sufficiently large $x$, we have $g''(x)>0$. Thus, an appropriate translation of $g(x)$--in particular, some function of the form $g(x+\alpha)$ for some sufficiently large $\alpha$--will give you a counterexample function $u$.
A: Consider the function
$$f(x):={1\over 1+x}\qquad(x>0)\ .$$
Then $0<f(x)<1$ and $f''(x)>0$ for all $x>0$.
